I tried to swap stdout and stderr and redirect each stream to the files, in bash interactive shell.

By using brace Grouping command, I gained the output what I want. But I do not understand why when I do not use Grouping command, the result is not be swapped.

$ # Using Grouping command. Swap operation is succeed.
$ { { echo stdout; echo stderr 1>&2; } 3>&1 1>&2 2>&3; } 1>1.txt 2>2.txt;
$ cat 1.txt
$ cat 2.txt

$ # Not using Grouping command. Swap is failed...
$ { echo stdout; echo stderr 1>&2; } 3>&1 1>&2 2>&3 1>1.txt 2>2.txt;
$ cat 1.txt
$ cat 2.txt

My understanding of the operation of swapping file descriptors with respect to the above command was as follows.

  1. 3>&1: redirect FD3 to FD1 (stdout)
  2. 1>&2: redirect FD1 to FD2 (stderr)
  3. 2>&3: redirect FD2 to FD3 (stdout)
  4. 1>1.txt: redirect FD1 to 1.txt (FD1 points to stderr)
  5. 2>2.txt: redirect FD2 to 2.txt (FD2 points to stdout)

But under my understanding, I think the result should not change depending on whether there are braces or not.

Am I making some basic misunderstandings? Why do the above two command results differ?

Info my env.

$ bash --version
GNU bash, version 4.4.12(1)-release (x86_64-apple-darwin16.3.0)
Copyright (C) 2016 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

Bash processes file descriptors in reverse order, so from right to left. First 2>2.txt, then 1>1.txt, and so on...

  • Are there any reference pages? I think this processing order is left to right. reference: Bash Reference Manual 3.6 Redirections > The following redirection operators may precede or appear anywhere within a simple command or may follow a command. Redirections are processed in the order they appear, from left to right. – fhiyo Oct 27 '17 at 1:08

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